h∩[g0,g0] = 0.
We next prove
Lemma 11 g0 is abelian.
Suppose that [g0,g0]6= 0. Since g0 is nilpotent, it has a non-zero center con-tained in [g0,g0]. Choose a non-zero element z∈[g0,g0] in this center. It can not be semi-simple for then it would lie inh. So it has a non-zero nilpotent part, n, which also must lie in the center ofg0, by theB ⊂Atheorem we proved in our section on linear algebra. But then adn adxis nilpotent for any x∈ g0 since [x, n] = 0. This implies thatκ(n,g0) = 0 which is impossible. QED Completion of proof of (5.3). We know thatg0 is abelian. But then, if h6=g0, we would find a non-zero nilpotent element ing0which commutes with all of g0 (proven to be commutative). Hence κ(n,g0) = 0 which is impossible.
This completes the proof of (5.3). QED So we have the decomposition
g=h⊕M
α6=0
gα
which shows that any maximal toral subalgebrahis aCSA.
Conversely, suppose thathis aCSA. For anyx=xs+xn∈g, g0(adxs)⊂ g0(adx) sincexnis an ad nilpotent element commuting with adxs. If we choose x ∈ h minimal so that h = g0(adx), we see that we may replace x by xs and write h =g0(adxs). But g0(adxs) contains some maximal toral algebra containingxs, which is then a Cartan subalgebra contained inhand hence must coincide with h. This completes the proof of the theorem. QED
5.5 Roots.
We have proved that the restriction ofκtohis non-degenerate. This allows us to associate to every linear functionφonhthe unique elementtφ∈hgiven by
φ(h) =κ(tφ, h).
The set ofα∈h∗, α6= 0 for whichgα6= 0 is called the set ofroots and is denoted by Φ. We have
• Φ spans h∗ for otherwise ∃h 6= 0 : α(h) = 0 ∀α ∈ Φ implying that [h,gα] = 0∀αso [h,g] = 0.
• α∈Φ⇒ −α∈Φ for otherwisegα⊥g.
• x∈gα, y∈g−α, α∈Φ⇒[x, y] =κ(x, y)tα. Indeed, κ(h,[x, y]) = κ([h, x], y)
= κ(tα, h)κ(x, y)
= κ(κ(x, y)tα, h).
• [gα,g−α] is one dimensional with basistα. This follows from the preceding and the fact that gα can not be perpendicular tog−α since otherwise it will be orthogonal to all of g.
• α(tα) =κ(tα, tα)6= 0. Otherwise, choosingx∈gα, y∈g−αwithκ(x, y) = 1, we get
[x, y] =tα,[tα, x] = [tα, y] = 0.
Sox, y, tαspan a solvable three dimensional algebra. Acting as ad on g, it is superdiagonizable, by Lie’s theorem, and hence adtα, which is in the commutator algebra of this subalgebra is nilpotent. Since it is ad semi-simple by definition ofh, it must lie in the center, which is impossible.
• Chooseeα∈gα, fα∈g−αwith
κ(eα, fα) = 2 κ(tα, tα). Set
hα:= 2 κ(tα, tα)tα.
Theneα, fα, hα span a subalgebra isomorphic tosl(2). Call itsl(2)α. We shall soon see that this notation is justified, i.e thatgαis one dimensional and hence that sl(2)α is well defined, independent of any “choices” of eα, fαbut depends only onα.
• Consider the action ofsl(2)αon the subalgebram:=h⊕Lgnαwheren∈ Z. The zero eigenvectors ofhαconsist ofh⊂m. One of these corresponds to the adjoint representation of sl(2)α ⊂ m. The orthocomplement of hα∈hgives dimh−1 trivial representations ofsl(2)α. This must exhaust all the even maximal weight representations, as we have accounted for all the zero weights of sl(2)α acting ong. In particular, dimgα = 1 and no integer multiple ofαother than−αis a root. Now consider the subalgebra p:=h⊕Lgcα, c∈C. This is a module for sl(2)α. Hence all suchc’s must be multiples of 1/2. But 1/2 can not occur, since the double of a root is not a root. Hence the ±α are the only multiples ofα which are roots.
Now consider β∈Φ, β6=±α. Let k:=M
gβ+jα.
Each non-zero summand is one dimensional, and k is ansl(2)α module. Also β+iα6= 0 for anyi, and evaluation on hαgivesβ(hα) + 2i. All weights differ by multiples of 2 and sokis irreducible. Let qbe the maximal integer so that β+qα∈ Φ, andr the maximal integer so that β−rα ∈Φ. Then the entire string
β−rα, β−(r−1)α, . . . β+qα
5.5. ROOTS. 83 are roots, and
β(hα)−2r=−(β(hα) + 2q) or
β(hα) =r−q∈Z.
These integers are called theCartan integers.
We can transfer the bilinear formκfromhtoh∗by defining (γ, δ) =κ(tγ, tδ).
So
β(hα) = κ(tβ, hα)
= 2κ(tβ, tα) κ(tα, tα)
= 2(β, α) (α, α).
So 2(β, α)
(α, α) =r−q∈Z.
Choose a basisα1, . . . , α` ofh∗ consisting of roots. This is possible because the roots spanh∗. Any rootβ can be written uniquely as linear combination
β=c1α1+· · ·+c`α`
where the ci are complex numbers. We claim that in fact the ci are rational numbers. Indeed, taking the scalar product relative to ( , ) of this equation with theαi gives the`equations
(β, αi) =c1(α1, αi) +· · ·+c`(α`, αi).
Multiplying the i-th equation by 2/(αi, αi) gives a set of `equations for the ` coefficientsciwhere all the coefficients are rational numbers as are the left hand sides. Solving these equations for theci shows that theci are rational.
LetE be therealvector space spanned by the α∈ Φ. Then (, ) restricts to a real scalar product onE. Also, for anyλ6= 0∈E,
(λ, λ) :=: κ(tλ, tλ) := tr(adtλ)2
= X
α∈Φ
α(tλ)2
> 0.
So the scalar product (, ) onE is positive definite. E is a Euclidean space.
In the string of roots,β isqsteps down from the top, soqsteps up from the bottom is also a root, so
β−(r−q)α
is a root, or
β−2(β, α) (α, α)α∈Φ.
But
β−2(β, α)
(α, α)α=sα(β)
wheresαdenotes Euclidean reflection in the hyperplane perpendicular toα. In other words, for everyα∈Φ
sα: Φ→Φ. (5.6)
The subgroup of the orthogonal group of E generated by these reflections is called the Weyl group and is denoted by W. We have thus associated to every semi-simple Lie algebra, and to every choice of Cartan subalgebra a finite subgroup of the orthogonal group generated by reflections. (This subgroup is finite, because all the generating reflections, sα, and hence the group they generate, preserve the finite set of all roots, which span the space.) Once we will have completed the proof of the conjugacy theorem for Cartan subalgebras of a semi-simple algebra, then we will know that the Weyl group is determined, up to isomorphism, by the semi-simple algebra, and does not depend on the choice of Cartan subalgebra.
We define
hβ, αi:= 2(β, α) (α, α). So
hβ, αi = β(hα) (5.7)
= r−q∈Z (5.8)
and
sα(β) =β− hβ, αiα. (5.9)
So far, we have defined the reflectionsαpurely in terms of the root struction on E, which is the real subspace of h∗ generated by the roots. But in fact, sα, and hence the entire Weyl group arises as (an) automorphism(s) ofgwhich preserveh. Indeed, we know thateα, fα, hαspan a subalgebrasl(2)αisomorphic tosl(2). Now exp adeα and exp ad(−fα) are elements ofE(g). Consider
τα:= (exp adeα)(exp ad(−fα))(exp adeα)∈ E(g). (5.10) We claim that
Proposition 16 The automorphismταpreserves hand onhit is given by
τα(h) =h−α(h)hα. (5.11)
In particular, the transformation induced by τα onE issα.
5.6. BASES. 85